Answer
$F\left( \theta \right) = - 2\cos \theta - 3\sec \theta + C$
Work Step by Step
$$\eqalign{
& f\left( \theta \right) = 2\sin \theta - 3\sec \theta \tan \theta \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}\sin x \to {\text{Particular antiderivative: }} - \cos x \cr
& {\text{Function: }}\sec x\tan x \to {\text{Particular antiderivative: }}\sec x \cr
& {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr
& {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr
& f\left( \theta \right) = 2\sin \theta - 3\sec \theta \tan \theta \cr
& F\left( \theta \right) = 2\left( { - \cos \theta } \right) - 3\left( {\sec \theta } \right) + C \cr
& {\text{Simplifying}} \cr
& F\left( \theta \right) = - 2\cos \theta - 3\sec \theta + C \cr
& \cr
& {\text{Checking the answer by differentiation}} \cr
& F'\left( \theta \right) = \frac{d}{{d\theta }}\left[ { - 2\cos \theta } \right] - \frac{d}{{d\theta }}\left[ {3\sec \theta } \right] + \frac{d}{{dx}}\left[ C \right] \cr
& F'\left( \theta \right) = - 2\left( { - \sin \theta } \right) - 3\left( {\sec \theta \tan \theta } \right) + 0 \cr
& F'\left( \theta \right) = 2\sin \theta - 3\sec \theta \tan \theta \cr} $$